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Chapter 7 Alkenes and Alkynes I: Properties and Synthesis Elimination Reactions of Alkyl Halides
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Slides from Chapter 7

Feb 03, 2017

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Page 1: Slides from Chapter 7

Chapter 7 Alkenes and Alkynes I:

Properties and Synthesis Elimination Reactions of Alkyl Halides"

Page 2: Slides from Chapter 7

The Cahn-Ingold-Prelog convention is used to assign the groups of highest priority on each carbon"

  If the group of highest priority on one carbon is on the same side as the group of highest priority on the other carbon the double bond is Z (zusammen)"

  If the highest priority groups are on opposite sides the alkene is E (entgegen)"

The (E)-(Z) System for Designating Alkene Diastereomers

Page 3: Slides from Chapter 7

Name this compound"

A.  E-2-chloro-3-methyl-2-pentene""B.  Z-2-chloro-3-methyl-2-pentene"

C.  Z-2-chloro-3-ethyl-2-butene"

D.  None of these"

C C

Cl

H3C

CH2CH3

CH3

Page 4: Slides from Chapter 7

This compound is"

A.  E diastereomer""B.  Z diastereomer"

C.  Chiral"

D.  Neither E nor Z"

CH3

Br

Page 5: Slides from Chapter 7

l  Generally cis alkenes are less stable than trans alkenes because of steric hinderance"

l  Heat of Hydrogenation" The relative stabilities of alkenes can be measured using the

exothermic heats of hydrogenation""

Relative Stabilities of Alkenes

Page 6: Slides from Chapter 7

Heats of hydrogenation of three butene isomers

Since product is the same, the difference in ∆H are due to the alkene

Page 7: Slides from Chapter 7

The greater the number of attached alkyl groups (i.e. the more highly substituted the carbon atoms of the double bond), the greater the alkeneʼs stability"

Overall Relative Stability of Alkenes

Page 8: Slides from Chapter 7

Dehydrohalogenation"l  Reactions by an E2 mechanism are most useful"

 E1 reactions can be problematic (not easy to control)"l  E2 reactions are favored by:"

 Secondary or tertiary alkyl halides" Alkoxide bases such as sodium ethoxide or potassium tert-

butoxide"l  Bulky bases such as potassium tert-butoxide should be

used for E2 reactions of primary alkyl halides"

Synthesis of Alkenes via Elimination Reactions

Page 9: Slides from Chapter 7

"""

"""

Zaitzevʼs Rule: when two different alkenes are possible in an elimination, the most highly substituted alkene will be the major product"

 This is true only if a small base such as ethoxide is used"

Zaitsev’s Rule: Formation of the Most Substituted Alkene

Some hydrogen halides can eliminate to give two different alkenes

Page 10: Slides from Chapter 7

t  T.S. reflects greater stability of the more substituted double bond"t  An example of Kinetic control : When several products are possible,

the one that is derived through the process having the lower activation energy is product that predominates."

Basis of Zaitsev’s Rule: Transition state has double bond character

Page 11: Slides from Chapter 7

Bulky bases such as potassium tert-butoxide have difficulty removing sterically hindered hydrogens and generally only react with more accessible hydrogens (e.g. primary hydrogens)."

""This usually leads to the formation of the less substituted alkene."

""

Formation of the Least Substituted Alkene Using a Bulky Base

Page 12: Slides from Chapter 7

All four atoms (H-C-C-L) must be in the same plane"

t  Anti coplanar orientation is preferred over syn coplanar orientation because all atoms are staggered in this conformation."

t  In a cyclohexane ring the eliminating substituents must be diaxial to be anti coplanar"

The Stereochemistry of E2 Reactions

NOTE:

The syn co-planar orientation does occur in some rigid systems »  Anti-coplanar orientation" »  Syn-coplanar orientation"

Page 13: Slides from Chapter 7

Neomenthyl chloride and menthyl chloride give different elimination products with sodium ethoxide"

"

Example of stereochemical requirements

Page 14: Slides from Chapter 7

In neomenthyl chloride, the chloride is in the axial position in the most stable conformation"

t  Two axial hydrogens are anti to the chlorine and can eliminate."t  The Zaitzev product predominates"

Page 15: Slides from Chapter 7

In menthyl chloride the molecule must first change to a less stable conformation to produce an axial chloride"l  Elimination is slower"l  Can yield only the less substituted (Hofmann) product

from anti elimination"

Anti elimination is not possible

Only 1 H is anti to the axial Cl

Page 16: Slides from Chapter 7

Primary alcohols are the most difficult to dehydrate; tertiary are the easiest."

""Recall that elimination is favored over substitution at higher temperatures."

Caveat: Rearrangements of the carbon skeleton can occur"

Acid Catalyzed Dehydration of Alcohols

Protonation by acids converts OH into a good leaving group +OH2

Typical acids used in dehydration are sulfuric acid and phosphoric acid (where the conjugate base is not a good nucleophile)

Page 17: Slides from Chapter 7

Mechanism for Dehydration of 2o and 3o Alcohols: E1

Step 1: protonation of the hydroxy group

Step 2: formation of a carbocation

Page 18: Slides from Chapter 7

Note that only a catalytic amount of acid is required since it is regenerated in the final step of the reaction.

Another molecule of the alcohol, a water molecule, or the conjugate base of the acid may accept the proton.

Step 3: carbocation loses a proton

Page 19: Slides from Chapter 7

Recall the stability of carbocations is:"

""""

The second step in which the carbocation forms is rate determining.""The transition state for this reaction has carbocation character. """Tertiary alcohols react the fastest because they produce the most stable tertiary carbocation-like transition state in the second step."

Carbocation Stability and the Transition State

Page 20: Slides from Chapter 7

Primary alcohols have a prohibitively large energy barrier"

The relative heights of ∆G‡ for the second step of E1 dehydration

3o 2o

1o

Page 21: Slides from Chapter 7

t  Primary alcohols cannot undergo E1 dehydration because of the instability of the carbocation-like transition state. "

t  In the E2 dehydration the first step is the same: protonation of the hydroxyl group to yield the good leaving group, water."

t  Unable to form a carbocation, the protonated alcohol waits until some weak base assists in an E2 reaction"

Mechanism for Dehydration of Primary Alcohols is E2

Page 22: Slides from Chapter 7

Rearrangements during Dehydration of Secondary Alcohols"l  Rearrangement of a carbocation occurs if a more stable carbocation

can form:"

"

The first two steps are to same as for any E1 dehydration"

Carbocation Stability and Molecular Rearrangements

+ H20

Page 23: Slides from Chapter 7

The less stable 2o carbocation rearranges by shift of a methyl group with its electrons to produce more stable 3o carbocation"l  This is called a 1,2 shift"""

The removal of a proton to form the alkene occurs to give the Zaitzev (most substituted) product as the major one:"

Page 24: Slides from Chapter 7

A hydride shift (migration of a hydrogen with its electrons) can also occur to yield the more stable carbocation:"

"

Carbocation rearrangements can lead to the expansion of rings:"

Page 25: Slides from Chapter 7

t  Alkynes can be obtained by two consecutive dehydrohalogenation reactions of a vicinal dihalide"

Synthesis of Alkynes by Elimination Reactions

Page 26: Slides from Chapter 7

1. By bromination and consecutive dehydrohalogenation reactions"

2. Geminal dihalides can also undergo consecutive dehydrohalogenation reactions to yield the alkyne"

Alkenes can be converted to alkynes

Page 27: Slides from Chapter 7

Recall that acetylenic hydrogens have a pKa of about 25 and are much more acidic than most other C-H bonds."

""

""""

l  The relative acidity of acetylenic hydrogens in solution is:"

"

"l  Acetylenic hydrogens can be deprotonated with relatively

strong bases (sodium amide is typical)"  The products are called alkynides"

The Acidity of Terminal Alkynes

Page 28: Slides from Chapter 7

t  Sodium alkynides can be used as nucleophiles in SN2 reactions"l  New carbon-carbon bonds are the result."l  Only primary alkyl halides can be used; otherwise, elimination

reactions predominate"

Substitutions on terminal alkynes

Page 29: Slides from Chapter 7

Generation of acetylene from carbon"

t  Big Bang® carbide cannons  -  at big savings!"

t  You probably remember these from days of your youth. "

t    Big Bang cannons have been in continuous production since 1912. "

t  We offer the most popular models at below factory prices! "

t  These are great gifts for the junior cannoneer (ages 10 to 100) fun for the whole family"Big Bang cannons are fired by putting water in

the barrel housing and then adding a measured amount of Bangsite (carbide powder).  The powder dissolves in the water creating acetylene gas.  The gas is then ignited by a spark....resulting in a safe and loud BOOM. 

Page 30: Slides from Chapter 7

t  Hydrogen adds to alkenes in the presence of noble metal catalysts."

t  Heterogeneous Catalysts: finely divided insoluble platinum, palladium or nickel catalysts"

t  Homogeneous Catalysts: the catalyst (typically rhodium or ruthenium based) is soluble in the reaction medium. "l  For example, Wilkinsonʼs catalyst is Rh[(C6H5)3P]3Cl"

t  This process is called a reduction or hydrogenation"l  An unsaturated compound becomes a saturated (with hydrogen)

compound:"

Hydrogenation of Alkenes

Page 31: Slides from Chapter 7

The catalyst provides a new reaction pathway with lower ΔG‡ values"

Hydrogenation: The Function of the Catalyst

Page 32: Slides from Chapter 7

"In heterogeneous catalysis the hydrogen and alkene adsorb onto the catalyst surface, and then a step-wise formation of C-H bonds occurs:"

""""

Both hydrogens add to the same face of the alkene (i.e., a syn addition)"

Page 33: Slides from Chapter 7

Asymmetric Hydrogenation"

Using chiral versions of Wilkinson’s catalyst, one can get stereoselectivity

For example, the asymmetric synthesis of L-dopa, a drug for treating Parkinson’s disease:

Page 34: Slides from Chapter 7

The reaction of hydrogen using regular metal catalysts results in formation of the alkane from an alkyne."

syn-Addition of Hydrogen: Synthesis of cis-Alkenes" The P-2 catalyst nickel boride results in the syn addition of one

equivalent of hydrogen to a triple bond." As a result, hydrogenation of an internal alkyne yields a cis alkene:"

Hydrogenation of Alkynes

Page 35: Slides from Chapter 7

l  Lindlarʼs catalyst also directs syn addition, stopping at the addition of only one molecule of hydrogen, thus producing cis-alkenes from alkynes "

Page 36: Slides from Chapter 7

Anti Addition of Hydrogen: Synthesis of trans-Alkenes""

A dissolving metal reaction which uses lithium or sodium metal at low temperature in ammonia or amine solvent produces trans-alkenes."

""Net anti addition occurs by the formal addition of hydrogen to the opposite faces of the triple bond"

Page 37: Slides from Chapter 7

Mechanism involves two sequential electron transfers from the metal

The vinylic anion prefers to be trans and this determines the trans stereochemistry of the product

Lithium donates an electron to the alkene (into its anti-bonding orbital!)

This produces a radical anion, which reacts as a base with the amine solvent:

Second electron converts the radical into an anion, which again acts as a base:

Page 38: Slides from Chapter 7

t  Saturated alkane = molecular formula CnH2n+2"t  Formula of CnH2n will have either a double bond or a ring"

t  A compound with general formula CnH2n-2 can have a triple bond, two double bonds, a double bond and a ring or two rings"

t  Index of Hydrogen Deficiency: the number of pairs of hydrogen atoms that must be subtracted from the molecular formula of the corresponding alkane to give the molecular formula of the compound under consideration."

Structural Information from Molecular Formulas

Page 39: Slides from Chapter 7

t  Example: A compound has the molecular formula C6H12"

t  Hydrogenation allows one to distinguish a compound with a double bond from one with a ring"

t  Compounds Containing Halogens, Oxygen, or Nitrogen"l  For compounds containing halogen atoms, the halogen atoms

are counted as if they were hydrogen atoms."l  Example: A compound with formula C4H6Cl2"

 This is equivalent to molecular formula C4H8 which has IHD=1"

Using the Index of Hydrogen Deficiency (IHD)_

Page 40: Slides from Chapter 7

t  For compounds containing oxygen, the oxygen is ignored and IHD is calculated based on the rest of the formula"l  Example: A compound with formula C4H8O has IHD = 1"

t  For compounds containing nitrogen, one hydrogen is subtracted for each nitrogen and the nitrogen is ignored in the calculation"l  Example: A compound with formula C4H9N is treated as if it has

formula C4H8 and has IHD = 1"

Page 41: Slides from Chapter 7

Benzene and Aromatic Rings"

t  Benzene has 3 double bonds and a ring "

t  IHD = 4"

t  Whenever IHD = 4 or more with a modest number of carbons, consider the possibility of an aromatic ring."

C6H6

Page 42: Slides from Chapter 7

What is the IHD of "

A) "1""B)" "2""C)" "3""D)" "4"

C6H8"

Page 43: Slides from Chapter 7

What is the IHD of "

A) "1""B)" "2""C)" "3""D)" "4"

C6H8BrON"

Page 44: Slides from Chapter 7

Retrosynthetic Analysis "

Organic synthesis is making complex molecules from simpler ones

Often we know where we want to finish (complex molecule) but not where to start so we work backwards

goal

Usually need to consider several options so we can optimize

Page 45: Slides from Chapter 7

Synthesis target – Vitamin B12"

R. B. Woodward, 1972 – 90 steps, 11 years, 100 chemists

Page 46: Slides from Chapter 7

Example: synthesis of 1-cyclohexyl-1-butyne"

After doing retrosynthetic analysis, we write the normal synthetic sequence and check each step for feasibility

Page 47: Slides from Chapter 7

Options for preparing 2-methylhexane: from compounds with fewer carbons "

Explore each of these further